exterior derivative

English

Noun

exterior derivative (plural exterior derivatives)

1. (calculus) A differential operator which acts on a differential k-form to yield a differential (k+1)-form, unless the k-form is a pseudoscalar, in which case it yields 0.

   The exterior derivative of a “scalar”, i.e., a function $f=f(x^{1},x^{2},...,x^{n})$ where the $x^{i}$’s are coordinates of $\mathbb {R} ^{n}$, is $df={\partial f \over \partial x^{1}}dx^{1}+{\partial f \over \partial x^{2}}dx^{2}+...+{\partial f \over \partial x^{n}}dx^{n}$.

   The exterior derivative of a k-blade $f\,dx^{i_{1}}\wedge dx^{i_{2}}\wedge ...\wedge dx^{i_{k}}$ is $df\wedge dx^{i_{1}}\wedge dx^{i_{2}}\wedge ...\wedge dx^{i_{k}}$.

   The exterior derivative $d$ may be though of as a differential operator del wedge: $\nabla \wedge$, where $\nabla ={\partial \over \partial x^{1}}dx^{1}+{\partial \over \partial x_{2}}dx^{2}+...+{\partial \over \partial x^{n}}dx^{n}$. Then the square of the exterior derivative is $d^{2}=\nabla \wedge \nabla \wedge =(\nabla \wedge \nabla )\wedge =0\wedge =0$ because the wedge product is alternating. (If u is a blade and f a scalar (function), then $fu\equiv f\wedge u$, so $d(fu)=\nabla \wedge (fu)=\nabla \wedge (f\wedge u)=(\nabla \wedge f)\wedge u=df\wedge u$.) Another way to show that $d^{2}=0$ is that partial derivatives commute and wedge products of 1-forms anti-commute (so when $d^{2}$ is applied to a blade then the distributed parts end up canceling to zero.)